180 HISTORY OF THE THEORY OF NUMBERS. [CHAP. IV 



and took r = t - - Ss, a = t 2 - mts + 61s 2 , and found t = 3839s/488, 

 r = 65s/488. J. Wright took a = r 2 + s 2 + ^rs and found r = Ss, 

 which does not give positive answers. Hence set r = t 8s. 



" Calculator " 98 found three right triangles of equal perimeters and 

 areas in arithmetical progression. The areas are proportional to the radii 

 r of the inscribed circles; for the sides 2amn, a(m 2 n 2 ), r = an(m ri). 

 A long computation yielded triangles all of whose sides involve eight digits : 



(18601944, 13951458, 23252430), (18559223, 13999464, 23247145), 



(18515584, 14048388, 23241860). 



W. Wright" found a right triangle whose perimeter is a square and 

 area a cube by taking ra 2 n 2 , 2mn as the sides. Let the perimeter equal 

 q 2 m 2 , whence m = 2n/(q 2 2). Then the area is a cube if 



Sn - 2n(q 2 - 2) 2 = s 3 , 



which gives n. '' Epsilon ' took p(m 2 n 2 ), 2pmn as the sides. The 

 perimeter is a square if p = 2m(m + n). The area is a cube if 4n(m n) 

 is, whence either n is a cube and the double of m n is a cube or vice versa. 



To find a right triangle the sum of whose sides equals the area, many 

 solvers 100 noted that 2s 2 + 2rs = rs(s 2 r 2 ) implies 2 = r 2 sr. The 

 root r involves the radical Vs 2 8, which is equated to s x, giving 

 s = (8 + x~)f(2x}. For integral solutions we have x < s, whence x = 4, 

 s = 3, r = 2 or 1 and the only triangles are (13, 5, 12), (10, 8, 6). 



J. Baines, 101 to find two right triangles the differences between whose 

 bases, perpendiculars, hypotenuses, perimeters and diameters of inscribed 

 circles are all squares, and difference of areas a cube, took 25m 2 n 2 , 

 Wmn and 25m 2 + n 2 as base, perpendicular and hypotenuse of one, and 

 9m 2 n 2 , 6mn, 9m 2 -f- n 2 for the other, so that we have only to make 4mn 

 and A = 32m 2 + 4mn squares and B = 98m 3 n 2mn 3 a cube. Take 

 mn = a 2 . Then A = D if 8a 2 + n 2 = D = (2ar/s + n) 2 . Take r = s = 1, 

 whence n = a = m. Then B = 96a 4 is a cube if a = & 3 /96. G. Heald 

 took the triangles (10z 2 , 24x 2 , 26z 2 ) and (6x 2 , 8x 2 , 10z 2 ). All but the last 

 condition is satisfied identically. The difference 96z 4 of the areas is a cube 

 if x = p 3 /12. 



J. Davey 102 found a right triangle whose perimeter is a square p 2 such 

 that p 3 equals the area. Take pr, ps, pt as the sides. Then r = p s t, 

 s = 2p/t, and r 2 = s 2 + t 2 , which gives p = 2t(t - - 2) /(t - - 4). 



Many 103 found the sides a, b and hypotenuse c of a right triangle such 

 that a, c + b, c b are integral cubes, say p 3 , m 3 , n 3 . Then c 2 b 2 = a 2 

 gives mn = p 2 . 



98 The Gentleman's Math. Companion, 4, No. 22, 1819, 861-4. Cf. Perkins. 104 

 "Ibid., 5, No. 28, 1825, 371-3. 



100 Ladies' Diary, 1828, 34, Quest. 1465. 



101 Ladies' Diary, 1830, 37, Quest. 1500. 



102 The Lady's and Gentleman's Diary, London, 1841, 58 (Quest. 1416 of Gentleman's Diary, 



1840). 



103 Ibid., 1845, 51-2, Quest. 1722. 



